How can I create Date Object from Date and Time Strings in Lua NodeMCU?

Question

I am playing around with NodeMCU on an ESP8266. I have a Date String and a Time String from a Web Request like this:

15.07.16 (German format DD.MM.YY)

19:50 (24 hours format)

These DateTimes lay usually a little bit in the future. I want to get the number of minutes from the current time to the time from my strings above.

I guess I have to create a time object from the strings and then compare it to the current time. But how can I do that with Lua?

Unfortunately there is no os Library on NodeMCU (or I might have missed how to enable it).

Calculating the difference manually would be a huge pain which I would like to avoid. Does anyone know a way to compute that with available or external libraries?

Thanks for any support!

Answer 1

There's a pending PR for rtctime that does the exact opposite, Unix epoch to UTC calendar.

If you convert your strings to a Unix epoch X you could do

-- delta in minutes
local delta = (X - rtctime.get()) / 60

You can either calculate X yourself, which is far from trivial due to leap years & seconds and other date/time oddities, or your can send a request to http://www.convert-unix-time.com/api?date=15.07.2016%2019:50&timezone=Vienna&format=german and extract the timestamp from it.

Answer 2

First you get the numbers from the strings using Lua's string library:

https://www.lua.org/pil/20.html

https://www.lua.org/manual/5.3/manual.html#6.4

Then you do the time calculations using Lua's os library:

https://www.lua.org/pil/22.1.html

https://www.lua.org/manual/5.3/manual.html#6.9

I won't give you more information as you did not show any own effort to solve the problem.

Addon:

As you don't have the os library (didn't know that) you can simply calculate that stuff yourself.

Get the month, year hour and minute number from the strings using string.sub or string patterns.

Then simply calculate the time difference. You know how many days each month has. You know how many minutes per hour and how many hours per day. Determine if the year is a leap year (if you don't know how: https://support.microsoft.com/en-us/kb/214019)